Question

1. $a^{3}b^{2}cdot(2b a^{-2})^{4}=$

Explanation:

Step1: Expand the power of the term

First, apply the power of a product rule \((xy)^n = x^n y^n\) and power rule \((x^m)^n = x^{mn}\) to \((2ba^{-2})^4\):
\[
(2ba^{-2})^4 = 2^4 \cdot b^4 \cdot (a^{-2})^4 = 16b^4a^{-8}
\]

Step2: Multiply with the first term

Now multiply \(a^3b^2\) with \(16b^4a^{-8}\), using the product rule \(x^m \cdot x^n = x^{m+n}\) for like bases:
\[
a^3b^2 \cdot 16b^4a^{-8} = 16 \cdot a^{3+(-8)} \cdot b^{2+4}
\]

Step3: Simplify the exponents

Calculate the sum of the exponents for each base:
\[
16 \cdot a^{-5} \cdot b^{6} = 16\frac{b^6}{a^5}
\]

Answer:

$\frac{16b^6}{a^5}$